Hi can I ask about the approach to this problem? Thanks!
# | User | Rating |
---|---|---|
1 | tourist | 3880 |
2 | jiangly | 3669 |
3 | ecnerwala | 3654 |
4 | Benq | 3627 |
5 | orzdevinwang | 3612 |
6 | Geothermal | 3569 |
6 | cnnfls_csy | 3569 |
8 | jqdai0815 | 3532 |
9 | Radewoosh | 3522 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | awoo | 161 |
2 | maomao90 | 160 |
3 | adamant | 156 |
4 | maroonrk | 153 |
5 | atcoder_official | 148 |
5 | -is-this-fft- | 148 |
5 | SecondThread | 148 |
8 | Petr | 147 |
9 | nor | 144 |
10 | TheScrasse | 142 |
Hi can I ask about the approach to this problem? Thanks!
Name |
---|
i dont know? thnks
bump
dumb
I believe there are many solutions to this problem. Anyways, here is my approach.
First, construct a "staircase" of size $$$m$$$ (full of 0s) on top of the grid then rotate everything 45 degrees counter-clockwise. Consider the sample :
the link for the picture in case you can't see it : https://ibb.co/pRTYZFF
If you construct another staircase of size $$$m$$$, full of 0s, at the bottom of the rotated grid, you can see that the red paralellogram is actually a rectangular submatrix of a $$$(n + m) * m$$$ matrix.
Any staircase can be represented as the difference of a green rectangle and a red "rectangle" as in the picture.
More specifically, let $$$A$$$ be the original grid and $$$B$$$ be the $$$(n + m) * m$$$ grid, the green rectangle is $$$(1, c) - (r, c + h - 1)$$$ in $$$A$$$, while the red rectangle is $$$(1, c) - (m + r - c - h, c + h - 1)$$$ in $$$B$$$. (here I'm denoting a rectangle by its top-left and bottom-right points)
We can use two 2D BITs to maintain $$$A$$$ and $$$B$$$ for the (sum of staircase) queries. For update queries $$$(r, c, v)$$$, update $$$[r][c] += v - A[r][c]$$$ in BIT $$$A$$$ and $$$[m + r - c][c] += v - B[m + r - c][c]$$$ in BIT $$$B$$$ and update the new values to $$$A[r][c]$$$ and $$$B[m + r - c][c]$$$.
Thanks you so much! I will look into it.